Kepler’s laws of Planetary Motion Questions in English

(C) From the given graph,the time period of rotation of the planet around the star is $T = 3 \text{ days}$.
According to Kepler's third law,the relationship between the orbital period $T$ and the orbital radius $r$ is given by:
$T^{2} = \frac{4 \pi^{2}}{G M} \cdot r^{3} \quad \dots(i)$
If we express the time period in years and the radius in $AU$,then for the Earth ($T = 1 \text{ yr}$,$r = 1 \text{ AU}$),the constant $\frac{4 \pi^{2}}{G M} = 1$.
Given $T = 3 \text{ days} = \frac{3}{365} \text{ yr}$,we substitute this into equation $(i)$:
$r^{3} = T^{2}$
$r = T^{2/3}$
$r = \left(\frac{3}{365}\right)^{2/3} \approx \left(\frac{1}{121.67}\right)^{2/3} \approx \left(0.0082\right)^{2/3} \approx 0.04 \text{ AU}$.
Thus,the radius of the planet's orbit is approximately $0.04 \text{ AU}$.

(A) By the law of conservation of angular momentum,the angular momentum $L$ of the planet remains constant at all points in its orbit.
At the perihelion $P$ and aphelion $A$,the velocity vector is perpendicular to the position vector,so the angular momentum is given by $L = mvr$.
Since $L_P = L_A$,we have $m v_P r_P = m v_A r_A$,which implies $\frac{v_P}{v_A} = \frac{r_A}{r_P}$.
For an ellipse with semi-major axis $a$ and eccentricity $e$,the distance of the closest point (perihelion) is $r_P = a(1-e)$ and the distance of the farthest point (aphelion) is $r_A = a(1+e)$.
Substituting these values,we get $\frac{v_P}{v_A} = \frac{a(1+e)}{a(1-e)} = \frac{1+e}{1-e}$.

(C) For a comet in an elliptical orbit,the semi-major axis $a$ is given by the average of the perihelion distance $(r_p)$ and the aphelion distance $(r_a)$:
$a = \frac{r_p + r_a}{2} = \frac{0.4 + x}{2} \, AU$
According to Kepler's Third Law,the square of the time period $T$ is proportional to the cube of the semi-major axis $a$:
$T^2 \propto a^3$
Comparing the comet with Earth ($T_e = 1 \, yr$,$a_e = 1 \, AU$):
$\left(\frac{T}{T_e}\right)^2 = \left(\frac{a}{a_e}\right)^3$
Substituting the given values $(T = 125 \, yr)$:
$\left(\frac{125}{1}\right)^2 = \left(\frac{0.4 + x}{2 \times 1}\right)^3$
$(125)^2 = \left(\frac{0.4 + x}{2}\right)^3$
Taking the cube root of both sides:
$(125)^{2/3} = \frac{0.4 + x}{2}$
$(5^3)^{2/3} = \frac{0.4 + x}{2}$
$5^2 = \frac{0.4 + x}{2}$
$25 = \frac{0.4 + x}{2}$
$50 = 0.4 + x$
$x = 50 - 0.4 = 49.6 \, AU$
Thus,the aphelion distance is $49.6 \, AU$.

(D) According to Kepler's third law of planetary motion,the square of the time period $T$ of a satellite is directly proportional to the cube of the radius $r$ of its orbit.
$T^2 \propto r^3$
This implies $T \propto r^{3/2}$ or $r \propto T^{2/3}$.
Since the exponent of $T$ is $2/3$ (which is less than $1$),the graph of $r$ versus $T$ will be concave downwards.
Looking at the provided graph,curve $(4)$ represents a relationship where $r$ increases with $T$ but with a decreasing slope (concave downwards),which matches the mathematical relationship $r \propto T^{2/3}$.
Therefore,the correct graph is $(4)$.

(A) The area $A$ of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is given by $A = \pi ab$.
For an elliptical orbit,the semi-minor axis $b$ is related to the semi-major axis $a$ and eccentricity $e$ by $b = a\sqrt{1-e^2}$.
Thus,$A = \pi a^2 \sqrt{1-e^2}$.
Since $e$ is a constant for a given orbit,$A \propto a^2$.
According to Kepler's $III$ law of planetary motion,the square of the time period $T$ is proportional to the cube of the semi-major axis $a$:
$T^2 \propto a^3$
$a \propto T^{2/3}$
Substituting this into the area relation:
$A \propto (T^{2/3})^2$
$A \propto T^{4/3}$
Therefore,the area of the elliptical orbit is proportional to $T^{4/3}$.

(D) The correct answer is $(D)$.
Kepler's first law,also known as the Law of Orbits,states that all planets move in elliptical orbits,with the sun situated at one of the two foci of the ellipse.

(C) According to Kepler's Third Law (Law of Periods),the square of the time period $T$ of revolution of a planet is proportional to the cube of the semi-major axis $a$ of the elliptical orbit.
$T^2 \propto a^3$
For an elliptical orbit with the sun at one focus,the maximum distance (aphelion) is $r_1 = a(1+e)$ and the minimum distance (perihelion) is $r_2 = a(1-e)$,where $e$ is the eccentricity.
Adding these two distances:
$r_1 + r_2 = a(1+e) + a(1-e) = 2a$
Therefore,the semi-major axis is:
$a = \frac{r_1 + r_2}{2}$
Substituting this into Kepler's Third Law:
$T^2 \propto \left(\frac{r_1 + r_2}{2}\right)^3$
Since $2^3$ is a constant,we have:
$T^2 \propto (r_1 + r_2)^3$
Taking the square root of both sides:
$T \propto (r_1 + r_2)^{3/2}$
Thus,the correct option is $C$.

(A) The gravitational force exerted by the Sun on a planet acts along the line joining the Sun and the planet.
Let $\vec{r}$ be the position vector of the planet with respect to the Sun and $\vec{F}_g$ be the gravitational force acting on the planet.
The torque $\vec{\tau}$ is given by $\vec{\tau} = \vec{r} \times \vec{F}_g$.
The magnitude of the torque is $\tau = r F_g \sin \theta$,where $\theta$ is the angle between the position vector $\vec{r}$ and the force vector $\vec{F}_g$.
Since the gravitational force is a central force,it acts along the line joining the centers of the two bodies. Therefore,the force vector $\vec{F}_g$ is directed towards the Sun,which is opposite to the position vector $\vec{r}$ (i.e.,$\theta = 180^{\circ}$).
Since $\sin(180^{\circ}) = 0$,the torque $\tau = 0$.

(A) The time period of a communication satellite is $T_c = 24 \, hrs$.
According to Kepler's third law of planetary motion,the square of the time period is directly proportional to the cube of the orbital radius,i.e.,$T^2 \propto r^3$.
Let $r_c$ be the radius of the communication satellite's orbit and $r_x$ be the radius of satellite $X$'s orbit. Given $r_x = \frac{1}{4} r_c$,or $\frac{r_c}{r_x} = 4$.
Using the ratio form of Kepler's law: $\frac{T_c}{T_x} = \left( \frac{r_c}{r_x} \right)^{3/2}$.
Substituting the values: $\frac{24}{T_x} = (4)^{3/2}$.
Since $(4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$,we have $\frac{24}{T_x} = 8$.
Therefore,$T_x = \frac{24}{8} = 3 \, hrs$.

(C) According to Kepler's third law of planetary motion,the square of the time period of revolution $(T)$ is directly proportional to the cube of the semi-major axis $(R)$ of the orbit: $T^2 \propto R^3$.
Given:
For Earth: $T_1 = 1 \, \text{year}$,$R_1 = 1.5 \times 10^8 \, \text{km}$.
For the imaginary planet: $T_2 = 2.83 \, \text{years}$,$R_2 = ?$.
Using the ratio formula:
$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3$
Substituting the values:
$\left(\frac{2.83}{1}\right)^2 = \left(\frac{R_2}{1.5 \times 10^8}\right)^3$
Since $2.83 \approx \sqrt{8}$,then $(2.83)^2 \approx 8$:
$8 = \left(\frac{R_2}{1.5 \times 10^8}\right)^3$
Taking the cube root on both sides:
$2 = \frac{R_2}{1.5 \times 10^8}$
$R_2 = 2 \times 1.5 \times 10^8 = 3 \times 10^8 \, \text{km}$.
Thus,the distance is $3 \times 10^8 \, \text{km}$.

(D) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the orbital radius $(R)$: $T^2 \propto R^3$.
Given the initial time period $T_1 = 24 \ hours$ and initial radius $R_1 = R$.
The new radius is $R_2 = \frac{R}{4}$.
Using the ratio formula: $\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$.
Substituting the values: $\left(\frac{24}{T_2}\right)^2 = \left(\frac{R}{R/4}\right)^3 = (4)^3 = 64$.
Taking the square root on both sides: $\frac{24}{T_2} = \sqrt{64} = 8$.
Therefore,$T_2 = \frac{24}{8} = 3 \ hours$.

(D) According to Kepler's second law of planetary motion,a planet moves faster when it is closer to the sun and slower when it is farther away.
Therefore,the linear speed of a planet revolving around the sun in an elliptical orbit is not constant.
Option $A$ is correct because the speed of a satellite in a circular orbit is given by $v = \sqrt{GM/r}$,which is constant for a fixed radius $r$.
Option $B$ is correct because the total mechanical energy of a planet in an elliptical orbit is conserved.
Option $C$ is correct because the mass of the earth is extremely large compared to any falling body,making its acceleration and displacement negligible.
Thus,the incorrect statement is $D$.

(A) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(r)$: $T^2 \propto r^3$.
Let the initial time period be $T_1 = 200 \text{ days}$ and the initial distance be $r_1 = r$.
Let the new time period be $T_2$ and the new distance be $r_2 = \frac{r}{4}$.
Using the relation $\frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3}$,we get:
$\frac{(200)^2}{r^3} = \frac{T_2^2}{(\frac{r}{4})^3}$
$T_2^2 = (200)^2 \times \frac{(\frac{r}{4})^3}{r^3}$
$T_2^2 = (200)^2 \times \frac{r^3}{64 \times r^3}$
$T_2^2 = \frac{(200)^2}{64}$
Taking the square root of both sides:
$T_2 = \frac{200}{\sqrt{64}}$
$T_2 = \frac{200}{8}$
$T_2 = 25 \text{ days}$.

(A) According to Kepler's Third Law of Planetary Motion, the square of the time period $(T)$ is proportional to the cube of the orbital radius $(R)$: $T^2 \propto R^3$.
Let $T_m$ and $R_m$ be the time period and orbital radius of the Moon, and $T_s$ and $R_s$ be the time period and orbital radius of the satellite.
Given: $R_s = R_m / 9$ and $T_m = 27 \text{ days}$.
Using the ratio formula: $\left(\frac{T_m}{T_s}\right)^2 = \left(\frac{R_m}{R_s}\right)^3$.
Substituting the values: $\left(\frac{27}{T_s}\right)^2 = \left(\frac{R_m}{R_m / 9}\right)^3 = (9)^3 = 729$.
Taking the square root on both sides: $\frac{27}{T_s} = \sqrt{729} = 27$.
Therefore, $T_s = \frac{27}{27} = 1 \text{ day}$.

(B) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$: $T^2 \propto R^3$.
Taking the logarithmic derivative of both sides,we get: $2 \frac{\Delta T}{T} = 3 \frac{\Delta R}{R}$.
Given the change in radius $\Delta R = 1.03 R - R = 0.03 R$,so $\frac{\Delta R}{R} = 0.03$.
Substituting this into the derivative equation: $2 \frac{\Delta T}{T} = 3 \times 0.03$.
Therefore,the percentage change in the time period is: $\frac{\Delta T}{T} \times 100 = \frac{3 \times 0.03}{2} \times 100 = 4.5 \%$.

(A) Kepler's second law states that the radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time. This implies that the areal velocity $\frac{dA}{dt}$ is constant.
The areal velocity is given by the formula $\frac{dA}{dt} = \frac{L}{2m}$,where $L$ is the angular momentum and $m$ is the mass of the planet.
Since the gravitational force exerted by the Sun on the planet is a central force,the torque $\tau$ acting on the planet about the Sun is zero $(\tau = \vec{r} \times \vec{F} = 0)$.
Because the torque is zero,the angular momentum $L$ remains constant over time.
Since $L$ and $m$ are constant,the areal velocity $\frac{dA}{dt}$ is also constant.
Therefore,Assertion $(A)$ is correct,and Reason $(R)$ provides the correct physical explanation for it.

(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the semi-major axis $r$ of the orbit: $T^2 \propto r^3$.
Given that the radius of the Martian orbit $r_M = 4 \times r_{Me}$,where $r_{Me}$ is the radius of the Mercury orbit.
The ratio of the time periods is given by: $\frac{T_{Me}}{T_M} = \left(\frac{r_{Me}}{r_M}\right)^{3/2}$.
Substituting the given values: $\frac{T_{Me}}{687} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$.
Therefore,$T_{Me} = \frac{687}{8} = 85.875$ days.
Rounding to the nearest whole number,the length of $1$ year on Mercury is approximately $88$ Earth days.

(A) According to Kepler's third law of planetary motion, the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis or the orbital radius $(r)$: $T^2 \propto r^3$.
Given that the initial time period $T_1 = 24 \text{ hours}$ and the initial radius is $r_1$.
The new radius is $r_2 = \frac{1}{4} r_1$.
Using the relation $\frac{T_2^2}{T_1^2} = \left( \frac{r_2}{r_1} \right)^3$, we get:
$\frac{T_2^2}{T_1^2} = \left( \frac{1}{4} \right)^3 = \frac{1}{64}$.
Taking the square root on both sides:
$\frac{T_2}{T_1} = \sqrt{\frac{1}{64}} = \frac{1}{8}$.
Therefore, $T_2 = \frac{T_1}{8} = \frac{24}{8} = 3 \text{ hours}$.

(A) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(r)$ of the orbit: $T^2 \propto r^3$.
This implies that $\frac{T_p^2}{T_e^2} = \frac{r_p^3}{r_e^3}$,where $T_p$ and $r_p$ are the time period and orbital radius of the planet,and $T_e$ and $r_e$ are those of the earth.
Given that $T_p = 8 T_e$,we substitute this into the ratio:
$\frac{(8 T_e)^2}{T_e^2} = \frac{r_p^3}{r_e^3}$
$64 = \frac{r_p^3}{r_e^3}$
Taking the cube root on both sides:
$\frac{r_p}{r_e} = (64)^{1/3} = 4$.
Therefore,the ratio of the radius of the planet's orbit to the radius of the earth's orbit is $4$.

(D) According to Kepler's Third Law of Planetary Motion,the square of the time period $(T)$ of a satellite is directly proportional to the cube of the radius $(r)$ of its orbit:
$T^2 \propto r^3$
Given for satellite $S_1$: radius = $r$,time period = $T_1$.
Given for satellite $S_2$: radius = $2r$,time period = $T_2$.
Using the proportionality:
$\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2}$
Substituting the values:
$\frac{T_2}{T_1} = \left( \frac{2r}{r} \right)^{3/2} = (2)^{3/2} = 2^{1} \cdot 2^{1/2} = 2\sqrt{2}$
Thus,the ratio is $2\sqrt{2}:1$.

(C) According to Kepler's Third Law of Planetary Motion, the square of the period of revolution $T$ is proportional to the cube of the semi-major axis $R$ of the orbit: $T^2 \propto R^3$.
Given that the period of planet $A$ $(T_A)$ is $8$ times the period of planet $B$ $(T_B)$, we have $T_A = 8 T_B$.
Using the relation $\frac{T_A}{T_B} = \left(\frac{R_A}{R_B}\right)^{3/2}$, we can write:
$\left(\frac{T_A}{T_B}\right)^2 = \left(\frac{R_A}{R_B}\right)^3$
Substituting the given values:
$(8)^2 = \left(\frac{R_A}{R_B}\right)^3$
$64 = \left(\frac{R_A}{R_B}\right)^3$
Taking the cube root on both sides:
$\frac{R_A}{R_B} = (64)^{1/3} = 4$
Thus, the distance of planet $A$ from the sun is $4$ times the distance of planet $B$ from the sun.

(B) According to Kepler's Third Law of Planetary Motion, the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis or the orbital radius $(r)$:
$T^2 \propto r^3$
Given the initial time period $T_1 = 5 \text{ hours}$ and the initial radius $r_1 = r$.
The new radius is $r_2 = 4r$.
Using the ratio formula:
$\frac{T_2^2}{T_1^2} = \left(\frac{r_2}{r_1}\right)^3$
Substituting the values:
$\frac{T_2^2}{T_1^2} = \left(\frac{4r}{r}\right)^3 = (4)^3 = 64$
Taking the square root on both sides:
$\frac{T_2}{T_1} = \sqrt{64} = 8$
Therefore, the new time period is:
$T_2 = 8 \times T_1 = 8 \times 5 \text{ hours} = 40 \text{ hours}$.

(B) According to Kepler's second law,the angular momentum of a planet revolving around the Sun is conserved because the gravitational force is a central force.
Therefore,$L_p = L_a$.
Perihelion is the point in the orbit closest to the Sun,and aphelion is the point farthest from the Sun.
Therefore,$r_p < r_a$.
Since angular momentum $L = mvr$ is conserved,we have $m v_p r_p = m v_a r_a$.
Since $r_p < r_a$,it follows that $v_p > v_a$.
Thus,the correct relation is $r_p < r_a, v_p > v_a, L_a = L_p$.

(D) When a planet revolves around the Sun,the gravitational force exerted by the Sun on the planet acts along the line joining them.
Since the torque $\vec{\tau} = \vec{r} \times \vec{F} = 0$,the angular momentum $\vec{L}$ of the planet about the Sun remains conserved.
According to Kepler's second law,the areal velocity (area swept per unit time) is proportional to the angular momentum $(dA/dt = L/2m)$.
Since the angular momentum $L$ and the mass $m$ of the planet are constant,the areal velocity remains constant.

(B) According to Kepler's second law of planetary motion,the areal velocity of a planet remains constant.
This means the area swept per unit time is constant:
$\frac{A_{1}}{t_{1}} = \frac{A_{2}}{t_{2}} = \frac{A_{3}}{t_{3}}$
Given $t_{1} = 2$ days,$t_{2} = 3$ days,and $t_{3} = 6$ days,we have:
$\frac{A_{1}}{2} = \frac{A_{2}}{3} = \frac{A_{3}}{6}$
To simplify this,multiply the entire equation by $6$:
$6 \times \frac{A_{1}}{2} = 6 \times \frac{A_{2}}{3} = 6 \times \frac{A_{3}}{6}$
$3 A_{1} = 2 A_{2} = A_{3}$

(C) According to Kepler's second law of planetary motion,the areal velocity of a planet remains constant. This implies that the angular momentum of the planet is conserved.
Since the angular momentum $L = mvr \sin(\theta)$ is constant,where $m$ is the mass of the planet,$v$ is its linear speed,$r$ is the distance from the sun,and $\theta$ is the angle between the position vector and the velocity vector.
At the perihelion (the point closest to the sun),the distance $r$ is minimum.
Therefore,to conserve angular momentum,the linear speed $v$ must be maximum at this point.
In the given diagram,point $A$ is the closest to the sun.
Thus,the linear speed of the planet is maximum at point $A$.

(A) According to Kepler's second law,the areal velocity of a planet is constant,which implies that the angular momentum $(L)$ of the planet remains constant.
The torque $(\tau)$ acting on a planet due to the gravitational force of the sun is given by $\tau = r \times F$. Since the gravitational force is a central force acting along the line joining the planet and the sun,the torque is zero $(\tau = 0)$.
From the relation $\tau = \frac{dL}{dt}$,if $\tau = 0$,then $\frac{dL}{dt} = 0$,which means $L$ is constant.
As the planet moves in an elliptical orbit,its distance from the sun changes,causing its linear speed,angular speed,and kinetic energy to vary with time.
Therefore,both Assertion $(A)$ and Reason $(R)$ are true,and Reason $(R)$ is the correct explanation for Assertion $(A)$.

(C) According to Kepler's second law of planetary motion,the radius vector joining a planet to the sun sweeps out equal areas in equal intervals of time.
That is,$\frac{dA}{dt} = \text{constant}$.
As shown in the figure,let $r$ be the position vector of the planet with respect to the sun and $F$ be the gravitational force on the planet due to the sun.
The torque $\tau$ exerted on the planet by this force about the sun is given by:
$\tau = r \times F = 0$
(Since $r$ and $F$ are collinear,i.e.,they act along the same line).
We know that torque is the rate of change of angular momentum:
$\tau = \frac{dL}{dt}$
Since $\tau = 0$,we have $\frac{dL}{dt} = 0$,which implies $L = \text{constant}$.
Thus,Kepler's second law is a direct consequence of the law of conservation of angular momentum.

(C) Let $T_A$ and $T_B$ be the time periods of planet $A$ and $B$ around the sun respectively. Given that $T_A = 8 T_B$.
According to Kepler's third law of planetary motion,the square of the time period is proportional to the cube of the semi-major axis (distance from the sun),i.e.,$T^2 \propto R^3$.
Therefore,$\left(\frac{T_A}{T_B}\right)^2 = \left(\frac{R_A}{R_B}\right)^3$.
Substituting the given values:
$\left(\frac{8 T_B}{T_B}\right)^2 = \left(\frac{R_A}{R_B}\right)^3$
$8^2 = \left(\frac{R_A}{R_B}\right)^3$
$64 = \left(\frac{R_A}{R_B}\right)^3$
Taking the cube root on both sides:
$\frac{R_A}{R_B} = (64)^{1/3} = 4$.
Thus,the distance of planet $A$ from the sun is $4$ times the distance of planet $B$ from the sun.

(C) According to Kepler's second law of planetary motion,the areal velocity of a planet or satellite moving in a central force field is constant.
Mathematically,the areal velocity is given by $\frac{dA}{dt} = \frac{L}{2m}$,where $L$ is the angular momentum and $m$ is the mass of the satellite.
Since $L = mvr \sin \theta$,we substitute this into the expression:
$\frac{dA}{dt} = \frac{mvr \sin \theta}{2m} = \frac{vr \sin \theta}{2}$.
As seen in the final expression,the mass $m$ cancels out.
Therefore,the areal velocity is independent of the mass of the satellite,meaning it is proportional to $m^0$.

(C) According to Kepler's second law,the areal velocity of a planet is constant. The time taken to sweep an area is proportional to the area swept.
Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Here $a = 2b$.
The Sun is at the focus $S(ae, 0)$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.
So,$S = (2 \cdot \frac{\sqrt{3}}{2}, 0) = (\sqrt{3}, 0)$.
The path $bed$ corresponds to the area swept by the radius vector from $S$ to the arc $bed$. The area of the ellipse is $A = \pi ab$.
The area swept by the radius vector in path $bed$ is $A_{bed} = \text{Area of rectangle} + \text{Area of sector}$.
Using the property of areal velocity,the time taken is proportional to the area swept. After calculating the areas for path $bed$ and $dab$,the ratio of times is found. Given $T_{bed} = 24$ hours = $1440$ minutes,the calculation yields $T_{dab} = 804$ minutes.

(B) The areal velocity $A$ is defined as the rate at which the area is swept by the position vector of the planet.
$A = \frac{dA}{dt} = \frac{1}{2} r^2 \omega$
Multiplying both sides by the mass $M$ of the planet:
$MA = \frac{1}{2} M r^2 \omega$
Since the moment of inertia $I$ of a point mass $M$ at distance $r$ is $I = M r^2$,we have:
$MA = \frac{1}{2} I \omega$
We know that the angular momentum $L$ is given by $L = I \omega$.
Substituting $L$ into the equation:
$MA = \frac{1}{2} L$
Therefore,the angular momentum $L$ is:
$L = 2MA$

(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(r)$ of the orbit: $T^2 \propto r^3$.
Let the initial orbit radius be $r_1$ and the initial time period be $T_1 = 24 \text{ hours}$ (since it is a geostationary satellite).
Let the new orbit radius be $r_2 = 2r_1$.
We need to find the new time period $T_2$.
Using the ratio: $\frac{T_2^2}{T_1^2} = \left( \frac{r_2}{r_1} \right)^3$.
Substituting the values: $\frac{T_2^2}{24^2} = \left( \frac{2r_1}{r_1} \right)^3 = 2^3 = 8$.
$T_2^2 = 8 \times 24^2$.
Taking the square root on both sides: $T_2 = \sqrt{8} \times 24 = 2\sqrt{2} \times 24 = 48\sqrt{2} \text{ hours}$.

(D) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$.
Let $T_1$ be the time period for radius $R_1 = R$ and $T_2$ be the time period for radius $R_2 = 1.02 R$.
Then,$\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2} = (1.02)^{3/2}$.
Using the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$,where $x = 0.02$ and $n = 1.5$:
$\frac{T_2}{T_1} \approx 1 + (1.5 \times 0.02) = 1 + 0.03 = 1.03$.
This implies $T_2 \approx 1.03 T_1$.
The percentage difference is given by $\frac{T_2 - T_1}{T_1} \times 100 = \left( \frac{T_2}{T_1} - 1 \right) \times 100$.
Substituting the value: $(1.03 - 1) \times 100 = 0.03 \times 100 = 3\%$.

(C) The areal velocity $\frac{d A}{d t}$ is given by the formula $\frac{d A}{d t} = \frac{1}{2} r^2 \omega$.
This formula is derived from the angular momentum $L = mvr = mr^2 \omega$,where the areal velocity $\frac{d A}{d t} = \frac{L}{2m} = \frac{1}{2} r^2 \omega$.
Since $\frac{1}{2}$ is a constant,we have $\frac{d A}{d t} \propto r^2 \omega$.
Therefore,the correct relation is $\frac{d A}{d t} \propto \omega r^2$.

(C) The semi-major axis $a$ of an elliptical orbit is the average of the maximum distance $(r_{max} = R)$ and minimum distance $(r_{min} = R/3)$:
$a = \frac{R + R/3}{2} = \frac{4R/3}{2} = \frac{2R}{3}$
According to Kepler's Third Law,the period of revolution $T$ is given by:
$T = 2\pi \sqrt{\frac{a^3}{GM}}$
Substituting the value of $a$:
$T = 2\pi \sqrt{\frac{(2R/3)^3}{GM}} = 2\pi \sqrt{\frac{8R^3}{27GM}}$
Squaring both sides:
$T^2 = 4\pi^2 \cdot \frac{8R^3}{27GM} = \frac{32\pi^2}{27GM} \cdot R^3$
Comparing this with $T^2 = \alpha R^3$,we get:
$\alpha = \frac{32\pi^2}{27GM}$

(A) According to Kepler's Second Law (Law of Areas),the radius vector joining the Sun and the planet sweeps out equal areas in equal intervals of time.
In the given figure,the Sun $(S)$ is at one of the foci of the elliptical orbit.
The area swept by the planet in path $DAB$ is the area of the region bounded by the arc $DAB$ and the lines $SD$ and $SB$.
The area swept by the planet in path $BCD$ is the area of the region bounded by the arc $BCD$ and the lines $SB$ and $SD$.
Since the Sun is closer to the side $A$,the area swept by the planet in path $DAB$ is smaller than the area swept in path $BCD$.
Therefore,the time taken to travel $DAB$ is less than the time taken to travel $BCD$.

(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $r$:
$T^2 \propto r^3$
Since angular speed $\omega = \frac{2\pi}{T}$,we have $T = \frac{2\pi}{\omega}$. Substituting this into the law:
$(\frac{2\pi}{\omega})^2 \propto r^3 \Rightarrow \frac{1}{\omega^2} \propto r^3 \Rightarrow r^3 \omega^2 = \text{constant}$.
For the earth $(E)$ and the planet $(P)$:
$r_E^3 \omega_E^2 = r_P^3 \omega_P^2$
Given $r_E = R$ and $\omega_P = 2\omega_E$:
$R^3 \omega_E^2 = r_P^3 (2\omega_E)^2$
$R^3 \omega_E^2 = r_P^3 (4\omega_E^2)$
$R^3 = 4 r_P^3$
$r_P^3 = \frac{R^3}{4} = \frac{R^3}{2^2}$
Taking the cube root on both sides:
$r_P = \frac{R}{2^{2/3}} = 2^{-2/3} R$.

(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ of a planet is directly proportional to the cube of its average distance $(R)$ from the Sun.
$T^{2} \propto R^{3}$
Let $T_{1} = D$ be the time period of the Earth at distance $L_{1}$,and $T_{2}$ be the time period of the other planet at distance $L_{2}$.
Then,$\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{L_{2}^{3}}{L_{1}^{3}}$
$\frac{T_{2}^{2}}{D^{2}} = \left(\frac{L_{2}}{L_{1}}\right)^{3}$
$T_{2}^{2} = D^{2} \left(\frac{L_{2}}{L_{1}}\right)^{3}$
Taking the square root on both sides:
$T_{2} = D \left(\frac{L_{2}}{L_{1}}\right)^{3/2} \text{ days}$.

(B) According to Kepler's Third Law,the time period $T$ of a planet orbiting a star of mass $M$ at a distance $R$ is given by $T^2 = \frac{4\pi^2 R^3}{GM}$.
Thus,$T \propto \sqrt{\frac{R^3}{M}}$.
For planet $P_1$: $T_1 \propto \sqrt{\frac{R^3}{2M}}$.
For planet $P_2$: $T_2 \propto \sqrt{\frac{(2R)^3}{4M}} = \sqrt{\frac{8R^3}{4M}} = \sqrt{\frac{2R^3}{M}}$.
Taking the ratio $\frac{T_2}{T_1}$:
$\frac{T_2}{T_1} = \frac{\sqrt{2R^3/M}}{\sqrt{R^3/2M}} = \sqrt{\frac{2R^3}{M} \cdot \frac{2M}{R^3}} = \sqrt{4} = 2$.
Therefore,the ratio of the time periods of revolution of $P_2$ and $P_1$ is $2$.